$(\mathbb{R}, \oplus, \odot)$ is a field? Given $a \oplus b=a+b+1$ and $a \odot b=a+b+ab$. We are given that $a \oplus b = a+b+1$ and $a \odot b= a+b+ab$ and I need to prove that it is a field. Do I just need to know that there is an inverse for $a \odot b$? So if I could let $a+b+ab=c$ then I want to find a $-c$ such that $c \odot -c=1$? Then the inverse, $-c=\frac{1}{a+b+ab}$ then? This would be that for every c I have a multiplicative inverse and since the set is a infinite I will not have any zero divisors and I have an integral domain. And since addition and multiplication is commutative here I also have a field.
Am I on the right track?
 A: Hint: Define $f:\Bbb R\to\Bbb R$ by $f(t)=t+1$. Then $$f(x\oplus y)=f(x)+f(y)$$and $$f(x\odot y)=f(x)f(y).$$(So $f$ is an isomorphism,  hence yes $(\Bbb R,\oplus,\odot)$ is a field.)
Detail: For the benefit of readers new to the notion of isomorphism: One might say one knows about field isomorphisms, but question how that can be relevant before we know that $(\Bbb R,\oplus,\odot)$ is a field.
That's a reasonable complaint. But the notion of "isomorphism" is more general than that - in general ann isomorphism between two structures is a bijection that preserves whatever operations and relations are relevant in the given context.
To be more explicit than people usually are, let's say an MAS ("Multiplication-Addition-Structure") is a set together with two binary operations of addition and multiplication. That's it, no further axioms. An MAS may or may not be a field. A bijection satisfying the two equations above is an MAS isomorphism, and it's obvious that if two MASs are isomorphic and one is a field then so is the other. The $f$ above is an MAS isomorphismm, qed.
To be even more explicit, the fact that $\oplus$ is commutative follows from the fact that $+$ is commutative and $f$ is an MAS isomorphism: $$a\oplus b=f^{-1}(f(a)+f(b))
=f^{-1}(f(b)+f(a))=b\oplus a.$$Similarly for all the other field axioms.
A: You need to verify all the field axioms.  The two operations are clearly commutative, but it is not obvious that $\odot$ is associative.  Then find the identities.  It is not true that $1$ is the identity for $\odot$, so your inverse for $c$ is not correct.  Finally you need to verify distributivity.
